A Question About Implicit Differentiation

[The Student]

In my grade 12 calculus course, there are endless examples of implicit differentiation, but none look like something like this: 2x^3 - x + y + y^2 = 10 where there is single y. And we need to find dy/dx. In this type of question, does the single y become ydy/dx or does it just become dy/dx? If the later is true, then this should be the answer: dy/dx = (-6x^2+1)/(1+2y) right? Again, if it is the later, then why doesn't chain rule apply when y only has a power of 1?

 

[mmm4444bot]

this should be the answer: dy/dx = (-6x^2+1)/(1+2y)

That is what I got. :cool:

I expressed it as \(\displaystyle \frac{dy}{dx} = -\frac{6x^2 - 1}{2y + 1}\)

why doesn't chain rule apply when y only has a power of 1?

I would say because y(x) is not a composite function in terms of itself and [y(x)]^2 is. I'm not satisfied with my wording, though.

 

[The Student]

That is what I got. :cool:

I expressed it as \(\displaystyle \frac{dy}{dx} = -\frac{6x^2 - 1}{2y + 1}\)




I would say because y(x) is not a composite function in terms of itself and [y(x)]^2 is. I'm not satisfied with my wording, though.

I am not sure what you mean here because why is a y^2 a composite function of itself and not y^1? What's the difference?

 

[mmm4444bot]

I knew that I blew it at the time I wrote it. The chain rule is being used when differentiating both y^2 and y, but I should not have expressed my perspective on why I see a difference between the two differentials because I cannot seem to adequately put it into words. So, please forget what I said about it in my previous reply.


To differentiate y^2, we use the Power Rule, and then the chain rule because y is a function of x.

2 * y^(2 - 1) * y´ = 2y * y´


To differentiate y^1, we use the Power Rule, and then the chain rule because y is a function of x.

1 * y^(1 - 1) * y´ = y^0 * y´


Is this explanation satisfactory? :cool:

 

[The Student]

I am not sure that I understand your question.

\(\displaystyle \dfrac{dy}{dy} = \displaystyle \lim_{\Delta y \rightarrow 0}\dfrac{\Delta y}{\Delta y} = \lim_{\Delta y \rightarrow 0}1 = 1.\)

\(\displaystyle \dfrac{dy}{dx} = 1 * \dfrac{dy}{dx} = \dfrac{dy}{dy} * \dfrac{dy}{dx}.\) Chain rule is alive and well.

Does this help?

Hmmm, let me put it this way.

d(y^2)/dx = 2y(dy/dx)

d(y)/dx = y(dy/dx) = y(1)


If the second interpretation of the derivitive of y using implicit differentiation is correct, then how does d(y)/dx = dy/dx and not just 1 or (y)dy/dx?

 

[srmichael]

Hmmm, let me put it this way.

d(y^2)/dx = 2y(dy/dx) -----> CORRECT

d(y)/dx = y(dy/dx) = y(1) -----> WRONG


If the second interpretation of the derivitive of y using implicit differentiation is correct, then how does d(y)/dx = dy/dx and not just 1 or (y)dy/dx?

The second interpretation of the derivative of y is NOT correct. I don't think you fully understood what JeffM stated in his reply above:

\(\displaystyle \dfrac{dy}{dx} = 1 * \dfrac{dy}{dx} = \dfrac{dy}{dy} * \dfrac{dy}{dx}\)

Now, your first interpretation of the derivative of y^2 is correct. The derivative of y^2 with respect to y is 2y and you nailed that. So....in the second interpretation the derivative of y with respect to y is just 1

 

[Deleted member 4993]

I am totally confused over the confusion of OP - I have not seen this before...

 

[mmm4444bot]

approaches math from his own perspective

frustrating when his perspective is not helpful

This is a nail head hit dead center.

OPINION: The Student's perspective is sometimes from within a tunnel, stubbornly refusing constructive criticism to come out. That's okay, though, because every student has the right to sit in a tunnel -- as long as they refrain from trying to place the responsibility of their decision onto the shoulders of those outside.

 

[The Student]

Yes, I totally get it now. Now I wonder how I was ever confused.

Thanks everyone!!!:grin: